Maximizing expected utility in Chinese Roulette requires Bayesian updating.
Let’s say on priors that P(n=1) = p and that P(n=5) = 1-p. Call this instance of the game G_p.
Let’s say that you shoot instead of quit the first round. For G_1/2, there are four possibilities:
n = 1, vase destroyed: The probability of this scenario is 1⁄12. No further choices are needed.
n = 5, vase destroyed. The probability of this scenario is 5⁄12. No further choices are needed.
n = 1, vase survived: The probability of this scenario is 5⁄12. The player needs a strategy to continue playing.
n = 5, vase survived. The probability of this scenario is 1⁄12. The player needs a strategy to continue playing.
Notice that the strategy must be the same for 3 and 4 since the observations are the same. Call this strategy S.
The expected utility, which we seek to maximize, is:
E[U(shoot and then S)] = 0 + 5⁄12 * (R + E[U(S) | n = 1]) + 1⁄12 * (R + E[U(S) | n = 5])
Most of our utility is determined by the n = 1 worlds.
Manipulating the equation we get:
E[U(shoot and then S)] = R/2 + 1⁄2 * (5/6 * E[U(S) | n = 1] + 1⁄6 * E[U(S) | n = 5])
But the expression 5⁄6 * E[U(S) | n = 1] + 1⁄6 * E[U(S) | n = 5] is the expected utility if we were playing G_5/6. So the optimal S is the optimal strategy for G_5/6. This is the same as doing a Bayesian update (1:1 * 5:1 = 5:1 = 5⁄6).
Maximizing expected utility in Chinese Roulette requires Bayesian updating.
Let’s say on priors that P(n=1) = p and that P(n=5) = 1-p. Call this instance of the game G_p.
Let’s say that you shoot instead of quit the first round. For G_1/2, there are four possibilities:
n = 1, vase destroyed: The probability of this scenario is 1⁄12. No further choices are needed.
n = 5, vase destroyed. The probability of this scenario is 5⁄12. No further choices are needed.
n = 1, vase survived: The probability of this scenario is 5⁄12. The player needs a strategy to continue playing.
n = 5, vase survived. The probability of this scenario is 1⁄12. The player needs a strategy to continue playing.
Notice that the strategy must be the same for 3 and 4 since the observations are the same. Call this strategy S.
The expected utility, which we seek to maximize, is:
E[U(shoot and then S)] = 0 + 5⁄12 * (R + E[U(S) | n = 1]) + 1⁄12 * (R + E[U(S) | n = 5])
Most of our utility is determined by the n = 1 worlds.
Manipulating the equation we get:
E[U(shoot and then S)] = R/2 + 1⁄2 * (5/6 * E[U(S) | n = 1] + 1⁄6 * E[U(S) | n = 5])
But the expression 5⁄6 * E[U(S) | n = 1] + 1⁄6 * E[U(S) | n = 5] is the expected utility if we were playing G_5/6. So the optimal S is the optimal strategy for G_5/6. This is the same as doing a Bayesian update (1:1 * 5:1 = 5:1 = 5⁄6).